The atomic number of Fluorine is 9
Valence (outer) electron configuration is : 2s²2p⁵
Therefore, it requires 1 electron in the p-orbital to complete its octet of 8 electrons.
Thus, the atom Fluorine generally will become <u>more </u>stable through the formation of an ionic chemical compound by accepting <u>1 </u> electron from another atom. This process will fill its outer energy level.
Ans: A) more, 1
<span>1) 0.2M ferric nitrate is added gradually to 1M sodium hydroxide. In result, a red precipitate appears. The precipitate is ferric hydroxide.
2) </span><span>0.2M potassium chromate is added gradually to 0.05M lead acetate. in result, a yellow precipitate appears. The precipitate is called potassium acetate.
The common between the two is that the colors originated from one of the reactants. </span>
The element that gains electrons, becomes reduced.
While the one which loses electrons, becomes oxidized.
In this equation,
CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.
By balancing the equation, we will get:
3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O
Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.
Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.
So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .
Given:
175 kilograms of Methane (CH4) to be synthesized into Hydrogen Cyanide (HCN)
The balanced chemical equation is shown below:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To calculate for the masses of ammonia and oxygen needed, our basis will be 175 kg CH4.
Molar mass:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
mass of NH3 = 185.94 kg NH3 needed
mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
mass of O2 = 525 kg
mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
mass of O = 131.25 kg O
